How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$

Question

Could it be possible to find the solution for the following series?

$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$

Thanks in advance!

Answer 1

HINT:

$$\sum_{n=0}^\infty a^n\sum_{m=0}^n b^m=\sum_{n=0}^\infty a^n \frac{1-b^{n+1}}{1-b}$$

Can you finish?

Answer 2

Since the OP admitted that they can't continue from Dr.MV's hint, I'll do it myself:

$$\sum_{n=0}^\infty a^n \frac{1-b^{n+1}}{1-b}=\frac{1}{1-b} \left(\sum_{n=0}^\infty a^n-b \sum_{n=0}^\infty (ab)^n \right)=$$

Here we use geometric series, so we need to have $|a|<1$ and $|ab|<1$: *

$$=\frac{1}{1-b} \left(\frac{1}{1-a}-\frac{b}{1-ab} \right)=\frac{1}{(1-a)(1-ab)}$$

* Correction due to Batominovski's comment

Answer 3

Let $$S = 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Also we know, $$(1-b)(1+b+b^2+b^3 \cdots +b^n-1) = 1-b^n$$
We now start feeling that having $(1-b)$ as a factor in each term in the $RHS$ will create a "nice" series.
So multiplying the $LHS$ & $RHS$ by $(1-b)$, we get,
      $$S(1-b) = (1-b)+a(1-b^2)+a^2(1-b^3)+a^3(1-b^4)+\cdots $$
Now, upon expanding the $RHS$, we get 2 geometric progressions,
     $$S(1-b) = (1+a+a^2+a^3+\cdots)-(b+ab^2+a^2b^3+a^3b^4+\cdots)$$
The first series is an infinite GP ( here $|a|<1$ is a constraint) with the first term equal to $1$ and common ratio $a$. So its sum, $S_1$ (say) is given by,
      $$S_1 = \frac{1}{1-a}$$
The second series is also an infinite GP ( here too $|ab|<1$ is a constraint). It's first term is $b$ and the common ratio is $ab$. It's sum, $S_2$ (say), is given by,
      $$S_2=\frac{b}{1-ab}$$
Combining $S_1$ and $S_2$ we finally get,
      $$S(1-b) = \frac{1}{1-a} - \frac{b}{1-ab}$$
Simplifying the $RHS$ and dividing both sides of the equation by $(1-b)$ we get,
                              $$S = \frac{1}{(1-a)(1-ab)}$$

Answer 4

A slight variant that requires no explicit cancelling of $1-b$ (so you needn't handle the $b=1$ case with a continuity argument) rearranges the double summation, viz. $$\sum_{n=0}^\infty\left(a^n\sum_{k=0}^n b^k\right)=\sum_{k=0}^\infty b^k\sum_{n=k}^\infty a^n=\sum_{k=0}^\infty b^ka^k\frac{1}{1-a}=\frac{1}{\left( 1-a\right)\left( 1-ab\right)}$$ provided $\left| a\right| < 1$ and $\left| ab\right| <1$.

Answer 5

You have $$\sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right)\text{.}$$ Before determining what this converges to, it is worth establishing for which values of $(a,b)$ it converges at all. To that end, consider the Ratio Test:

$$\left\lvert\frac{a^{n+1}\sum_{m=0}^{n+1}b^m}{a^n\sum_{m=0}^nb^m}\right\rvert=\left\lvert a\left(1+\frac{b^{n+1}}{\sum_{m=0}^nb^m}\right)\right\rvert\to\begin{cases}\lvert ab\rvert&\lvert b\rvert>1\\\lvert a\rvert&\lvert b\rvert=1\\\lvert a\rvert&\lvert b\rvert<1\\\end{cases}$$

So it converges when:

• $\lvert ab\rvert<1$ and $\lvert b\rvert>1$
• $\lvert a\rvert<1$ and $\lvert b\rvert\leq1$

And diverges when:

• $\lvert ab\rvert>1$ and $\lvert b\rvert>1$
• $\lvert a\rvert>1$ and $\lvert b\rvert\leq1$

This test has been inconclusive when:

• $\lvert ab\rvert=1$ and $\lvert b\rvert>1$
• $\lvert a\rvert=1$ and $\lvert b\rvert\leq1$

Let's look at the last case first. If $\lvert a\rvert=1$, then you have $\sum_{n=0}^\infty {\pm_n\left(\sum_{m=0}^nb^m\right)}\text{.}$ The terms of such a series, $\pm_n\left(\sum_{m=0}^nb^m\right)$, do not approach $0$ no matter what $b$ is. So your series will diverge.

What if $\lvert b\rvert>1$ and $\lvert a\rvert=\frac{1}{\lvert b\rvert}$? Since $\lvert b\rvert>1$ we may write $\sum_{m=0}^nb^m=\frac{b^{n+1}-1}{b-1}$, and your series is $\sum_{n=0}^\infty {\pm_n\left(\frac{b^{n+1}-1}{b^n(b-1)}\right)}\text{.}$ Again using $\lvert b\rvert>1$, the terms of this series do not converge to $0$, so this series would be divergent.

Now we know the only situations where your series converges are:

• $\lvert ab\rvert<1$ and $\lvert b\rvert>1$ (which implies $\lvert a\rvert<1$)
• $\lvert a\rvert<1$ and $\lvert b\rvert\leq1$ (which implies $\lvert ab\rvert<1$)

When $b\neq1$, both situations give: $$\begin{align} \sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right) &=\sum_{n=0}^\infty a^n\frac{1-b^{n+1}}{1-b}\\ &=\frac{1}{1-b}\left(\sum_{n=0}^\infty a^n-b\sum_{n=0}^\infty (ab)^n\right)\\ &=\frac{1}{1-b}\left(\frac{1}{1-a}-\frac{b}{1-ab}\right)\\ &=\frac{1}{(1-a)(1-ab)}\\ \end{align}$$

And when $b=1$ with $\lvert a\rvert<1$, $$\begin{align} \sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right) &=\sum_{n=0}^\infty (n+1)a^n\\ &=\left.\sum_{n=0}^\infty \frac{d}{dx}x^{n+1}\right|_{x=a}\\ &=\left.\frac{d}{dx}\sum_{n=0}^\infty x^{n+1}\right|_{x=a}\\ &=\left.\frac{d}{dx}\left(\frac{1}{1-x}-1\right)\right|_{x=a}\\ &=\left.\frac{1}{(1-x)^2}\right|_{x=a}\\ &=\frac{1}{(1-a)^2}\\ \end{align}$$

Which conveniently agrees with the formula $\frac{1}{(1-a)(1-ab)}$.